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An electron of mass m when accelerated through a potential difference V has de Broglie wavelength λ. The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be
- a)λ m/M
- b)λ √m/M
- c)λ M/m
- d)λ √M/m
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
An electron of mass m when accelerated through a potential difference ...
Explanation:
This problem can be solved using the de Broglie wavelength formula:
\[\lambda = \frac{h}{\sqrt{2mE}}\]
where h is the Planck constant, m is the mass of the particle, and E is the energy of the particle.
Given:
- For an electron: \[\lambda_{e} = \frac{h}{\sqrt{2m_{e}eV}}\]
- For a proton: \[\lambda_{p} = \frac{h}{\sqrt{2m_{p}eV}}\]
where m_e is the mass of the electron, m_p is the mass of the proton, V is the potential difference, and e is the charge of an electron.
Comparing the two wavelengths:
\[\frac{\lambda_{p}}{\lambda_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}\]
\[\lambda_{p} = \lambda_{e} \times \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}\]
Therefore, the de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be \[\lambda = \lambda_{e} \times \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}\]
This can be simplified as \[\lambda = \lambda_{e} \times \sqrt{\frac{m_{e}}{m_{p}}}\]
Hence, the correct answer is option B: \[\lambda = \lambda_{e} \times \sqrt{\frac{m_{e}}{m_{p}}}\]
This problem can be solved using the de Broglie wavelength formula:
\[\lambda = \frac{h}{\sqrt{2mE}}\]
where h is the Planck constant, m is the mass of the particle, and E is the energy of the particle.
Given:
- For an electron: \[\lambda_{e} = \frac{h}{\sqrt{2m_{e}eV}}\]
- For a proton: \[\lambda_{p} = \frac{h}{\sqrt{2m_{p}eV}}\]
where m_e is the mass of the electron, m_p is the mass of the proton, V is the potential difference, and e is the charge of an electron.
Comparing the two wavelengths:
\[\frac{\lambda_{p}}{\lambda_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}\]
\[\lambda_{p} = \lambda_{e} \times \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}\]
Therefore, the de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be \[\lambda = \lambda_{e} \times \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}\]
This can be simplified as \[\lambda = \lambda_{e} \times \sqrt{\frac{m_{e}}{m_{p}}}\]
Hence, the correct answer is option B: \[\lambda = \lambda_{e} \times \sqrt{\frac{m_{e}}{m_{p}}}\]
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Community Answer
An electron of mass m when accelerated through a potential difference ...
Explanation:
Given:
- Mass of electron (m) accelerated through a potential difference V has de Broglie wavelength λ.
- Mass of proton (M) accelerated through the same potential difference V is to be determined.
De Broglie wavelength formula:
λ = h / p
where h is the Planck's constant and p is the momentum of the particle.
Relation between momentum and potential difference:
For an accelerated particle, momentum can be related to the potential difference by the equation:
p = √(2mV)
De Broglie wavelength for electron:
Substitute the momentum of the electron in the de Broglie wavelength formula:
λ = h / √(2mV)
De Broglie wavelength for proton:
Now, for the proton accelerated through the same potential difference V:
p = √(2MV)
Substitute the momentum of the proton in the de Broglie wavelength formula:
λ' = h / √(2MV)
Comparing the wavelengths:
To compare the wavelengths of the electron and proton:
λ' / λ = (h / √(2MV)) / (h / √(2mV))
λ' / λ = √(m / M)
Therefore, the de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:
λ' = λ √(m / M) (Option 'B')
Given:
- Mass of electron (m) accelerated through a potential difference V has de Broglie wavelength λ.
- Mass of proton (M) accelerated through the same potential difference V is to be determined.
De Broglie wavelength formula:
λ = h / p
where h is the Planck's constant and p is the momentum of the particle.
Relation between momentum and potential difference:
For an accelerated particle, momentum can be related to the potential difference by the equation:
p = √(2mV)
De Broglie wavelength for electron:
Substitute the momentum of the electron in the de Broglie wavelength formula:
λ = h / √(2mV)
De Broglie wavelength for proton:
Now, for the proton accelerated through the same potential difference V:
p = √(2MV)
Substitute the momentum of the proton in the de Broglie wavelength formula:
λ' = h / √(2MV)
Comparing the wavelengths:
To compare the wavelengths of the electron and proton:
λ' / λ = (h / √(2MV)) / (h / √(2mV))
λ' / λ = √(m / M)
Therefore, the de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:
λ' = λ √(m / M) (Option 'B')
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An electron of mass m when accelerated through a potential difference V has de Broglie wavelength λ. The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will bea)λ m/Mb)λ √m/Mc)λ M/md)λ √M/mCorrect answer is option 'B'. Can you explain this answer?
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